Transcript Class3

EE130
Electromechanics
2013
J. Arthur Wagner, Ph.D.
Prof. Emeritus in EE
[email protected]
Fig. 2.1
Electric Drive
System
(ASD)
Example of
load-speed
requirement
Load
ASD
L
 L  rad /sec 
desired speed profile
100
0
1
2
3
4
5
6
7 t  sec 
Fig. 2.2 (Linear) motion of M
fe
M
fL

fM
M
x
dx
u
;
dt
fe  f L
du
a

dt
M
dx
u
;
dt
du
fM
a

dt
M
Write the formula for
acceleration (sum of forces /
mass) (2.2)
Write the formula for power
(Net force times velocity) (2.6)
Write the formula for kinetic
energy (Recall from physics)
(2.9)
Accelerati
on, Power
Input,
Kinetic
Energy
Fig. 2.3 (a) Pivoted lever (b)
Holding torque for the lever
f
f
90 o

M
r

torque
Torque = force x radius

Mg
Ex. 2.1
• M = 0.5 kg
• r = 0.3 m
• Calculate holding torque as a function of
beta
• How do we work this?
Ex. 2.1
•
•
•
•
•
•
•
torque = force x radius
force perp. to radius
force = Mg
perp. component = Mg cos (beta)
torque = 0.5 kg 9.8 m/s^2 * 0.3 m cos(beta)
= 1.47 Nm
In a motor, the force is produced
electromagnetically and is tangent to the
cylindrical rotor. The rotor radius converts this
force to torque on the shaft.
Fig. 2.4 Motor torque acting on
an inertia load

Motor
Tem
TL
Load
the inertia of a cylinder = ½ * M * r1^2 (2.20)
M = cylinder mass
r1 = cylinder radius
TL = load torque, other than due to inertia
An inertia Load
• Mostly inertia
• Mostly friction (a grinder)
• Mostly mechanical torque in steady state (a
belt lifting gravel)
• All systems have some of both (an inertia
plus other torque)
Inertia of a 14 in disk
•
•
•
•
•
14.5 oz /(16 oz/lb) / (2.2 lb / kg) = .412 kg
d1 = 14 in * (.0254 m / in) = .356 m
r1 = .356 / 2 = .178 m
J = ½ * M * r1^2 = ½ *.412 * (.178)^2
= 0.0065 kg m^2
Angular acceleration = torque /
moment of inertia (2.23)
Tem  TL  TJ
d Tem  TL TJ


dt
J
J
Fig. 2.6 Motor and load with
rigid coupling
m
Motor
Tem


TL
TJ
1 
J eq
TL
Tem

m


Load
Block diagram of
acceleration equation.
Two integrators to go to
angular velocity and
angular position
Ex. 2.3
• TL negligible, each cylinder has same
inertia
• J of one cylinder = .029 kg m^2
• speed goes from 0 to 1800 rpm in 5 s
• Calculate the required electromagnetic
torque.
• How do we work this? First, sketch a speed
profile.
Ex. 2.3
• 1800 rpm * pi / 30 = 188.5 rad/s
• Jeq = 2 * .029 = 0.058 kg m^2
• acceleration = 188.5 rad/s / 5 s = 37.7
rad/s^2
• electromagnetic torque = J * accel =
• =.058 * 37.7 = 2.19 N m
Fig. 2.13 Combination of rotary
and linear motion J  motor inertia
m
M = mass of load
fL
M
r
Tem
Motor
Jm
u
r = pulley radius
Rotary and Linear Motion
du
f  M
 fL
dt
u  r m
2 d m
T  r f r M
 rf L
dt
Tem 
d m
Jm
dt
required to accelerate
motor
T due to load only
d
r M
 r fL
dt
2
due to load
Homework Chapter 2, Due next
Tuesday
• Problems 2.1, 2.11, 2.12, 2.13, 2.14